The number of zeros at the end of a number is determined by how many times the number can be divided by 10, which is the product of 2 and 5. Therefore, the number of 2s and 5s in the factorization of the expression will determine the number of trailing zeros, with the smaller of the two counts being the final result.

Consider the expression: $1 \times 5 \times 10 \times 15 \times 20 \times 25 \times 30 \times 35 \times 40 \times 45 \times 50 \times 55 \times 60$

Breaking it down into prime factors: $1 \times 5 \times (5 \times 2) \times (5 \times 3) \times (5 \times 2^2) \times (5 \times 5) \times (5 \times 3 \times 2) \times (5 \times 7) \times (5 \times 2^3) \times (5 \times 9) \times (5 \times 5 \times 2) \times (5 \times 11) \times (5 \times 2^2 \times 3)$

Here, the number of 2s is 10, and the number of 5s is 14. Since the number of zeros at the end is determined by the smaller of these two counts, there will be 10 zeros at the end of the expression.